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Standard Deviation

  1. Mar 19, 2006 #1

    Hootenanny

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    Just a quick question as I'm writing up my coursework. I'm calculating the standard deviation of a sample, my data is rate of reaction defined as [itex]\frac{1}{time} \times 10^{-3}[/itex]. So my input data is say 0.847, but the absolute value is 0.000847. The standard deviation formula returned a value of about 0.6. Is this my standard deviation or is it actually 0.0006? I'm inclined towards 0.6, but I'm not sure. Any help would be appreciated.
     
  2. jcsd
  3. Mar 19, 2006 #2
    f=1/T - sample frequency is in miliseconds?

    You made reading of 0.847, so thats 0.847ms? I think the deviation would be in milliseconds also, so 0.6ms or 0.0006s?

    or am i way off :?
     
  4. Mar 19, 2006 #3

    Hootenanny

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    This is a biology pratical. I was timing how long it took for a reaction to occur. For example, if the reaction took 19 minutes 40 seconds, that is a reaction time of 1180 seconds. As [itex]rate = \frac{1}{time}[/itex], the rate of reaction would be;
    [tex]rate = \frac{1}{1180} = 8.47\times 10^{-4} = 0.847\times 10^{-3} = 0.000847[/tex]

    I apologise if I mislead you with the information.
     
  5. Mar 19, 2006 #4

    Hootenanny

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    'bump' anybody else have an opinion?
     
  6. Mar 19, 2006 #5
    Do you have a list of data points?
     
  7. Mar 19, 2006 #6

    Hootenanny

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    Yes, do you want me to post them? Or attach them as an Excel spreadsheet? I could post them in latex if you like.
     
  8. Mar 19, 2006 #7
    Please do. Just make them standard text so I can just copy and paste it into excel. No need to tex all that.
     
    Last edited: Mar 19, 2006
  9. Mar 19, 2006 #8

    Hootenanny

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    Okay, I've got a tool for converting excel into Latex though :wink:

    --------------
    data
    --------------
    Temperature (°C) Time (secs) Rate (1/sec) x10-3

    17 1180 0.847
    1179 0.848
    1169 0.855
    1217 0.822
    Mean 1186.25 0.843
    32 180 5.556
    182 5.495
    180 5.556
    178 5.618
    Mean 180 5.556
    34.5 175 5.714
    179 5.587
    181 5.525
    210 4.762
    Mean 186.25 5.369
    39.5 100 10.000
    116 8.621
    116 8.621
    120 8.333
    Mean 113 8.850
    48 90 11.111
    90 11.111
    91 10.989
    93 10.753
    Mean 91 10.989
    50 105 9.524
    104 9.615
    104 9.615
    109 9.174
    Mean 105.5 9.479
    57 154 6.494
    143 6.993
    133 7.519
    125 8.000
    Mean 138.75 7.207
    64 1977 0.506
    2220 0.450
    2221 0.450
    2210 0.452
    Mean 2157 0.464
     
  10. Mar 19, 2006 #9

    Hootenanny

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    Uhh, that doesn't look right :S
     
  11. Mar 19, 2006 #10
    Ok, I need some help understanding your data here. You seems you have 8 runs each of size 4. What do you want to calculate the standard deviation for? The means? For all the data?
     
  12. Mar 19, 2006 #11

    Hootenanny

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    No, I calculated the standard deviation of each run. I know its only a small data set, so i used sample SD. Is that valid?
     
  13. Mar 19, 2006 #12
    Use the following:

    [tex]s = \sqrt { \frac{1}{n-1}( \sum^n_{i=1}X^2_i - n \bar{X}^2} )[/tex]

    Yes, you used the right one, because you are taking data from a sample of the population, your data is not for the entire population. It does not matter how big your data set is, if its not the actual population it is still going to be a sample standard deviation.
     
    Last edited: Mar 19, 2006
  14. Mar 19, 2006 #13

    Hootenanny

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    Yes I have done. Can I just ask you what you thing of my origonal question in post #1?
     
  15. Mar 19, 2006 #14
    I'm sorry but I don't understand your origional question. Can you rephrase it? You dont have an 'actual' standard deviation, because you are sampling data. The only way you can have an acutal standard deviation is if you had all the population data.
     
  16. Mar 19, 2006 #15

    Hootenanny

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    I mean I inputed the data direcly from my table (using excel), but my units for rate is [itex]\times 10^{-3}[/itex]. Say if the formula returned a SD value of 0.6, would it actually be 0.0006 due to the [itex]\times 10^{-3}[/itex]? I'm sorry if I'm not very lucid, but I can't think of another way of explaining it.
     
  17. Mar 19, 2006 #16
    well, then in that case:

    [tex] \sigma_{aX+b} =|a| \sigma_X [/tex]

    so multiply it by the absolute value of [itex]10^{-3} [/itex]
     
  18. Mar 19, 2006 #17

    Hootenanny

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    Ahhh, thank-you very much cyrus.
     
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